# Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 3

### Question 30(i). If 0 ≤ x≤ π and x lies in the 2nd quadrant such that sinx = 1/4, Find the values of cos(x/2), sin(x/2), and tan(x/2).

**Solution:**

Given that,

sinx = 1/4

As we know that, sinx = √(1 – cos

^{2}x)So,

⇒ (1/4)

^{2}= (1 – cos^{2}x)⇒ (1/16) – 1 = – cos

^{2}xcosx = ± √15/4

It is given that x is in 2nd quadrant, so cosx is negative.

cosx = – √15/4

Now,

As we know that, cosx = 2 cos

^{2}(x/2) – 1So,

⇒ – √15/4 = 2cos

^{2}(x/2) – 1⇒ cos

^{2}(x/2) = – √15/8 + 1/2cos(x/2) = ± (4-√15)/8

It is given that, x is in 2nd quadrant, so cos(x/2) is positive.

cos(x/2) = (4 – √15)/8

Again,

cosx = cos

^{2}(x/2) – sin^{2}(x/2)⇒ – √15/4 = {(4 – √15)/8}

^{2}– sin^{2}(x/2)⇒ sin

^{2}(x/2) = (4 + √15)/8⇒ sin(x/2) = ± √{(4 + √15)/8} = √{(4 + √15)/8}

Now,

tan(x/2) = sin(x/2) / cos(x/2)

=

=

=

=

=

= 4 + √15

Hence, the value of cos(x/2) = (4 – √15)/8, sin(x/2) = √{(4 + √15)/8}, and tan(x/2) = 4 + √15 .

### Question 30(ii). If cosx = 4/5 and x is acute, find tan2x.

**Solution:**

Given that,

cosx = 4/5

As we know that, sinx = √(1 – cos

^{2}x)So,

= √(1 – (4/5)

^{2})= √(1 – 16/25)

= √{(25 – 16)/25}

= √(9/25)

= 3/5

Since, tanx = sinx/cosx, so

= (3/5) / (4/5)

= 3/4

As we know that,

tan2x = 2tanx / (1 – tan

^{2}x)= 2(3/4) / {1 – (3/4)

^{2}}= 2(3/4) / (1 – 9/16)

= (3/2) / (7/16)

= 24/7

Hence, the value of tan2x is 24/7

### Question 30(iii). If sinx = 4/5 and 0 < x < π/2, then find the value of sin4x.

**Solution:**

Given that,

sinx = 4/5

As we know that, sinx = √(1 – cos

^{2}x)So,

⇒ (4/5)

^{2}= 1 – cos^{2}x⇒ 16/25 – 1 = -cos

^{2}x⇒ 9/25 = cos

^{2}x⇒ cosx = ±3/5

It is given that, x is ln the 1st quadrant

So, cosx = 3/5

Now,

sin4x = 2 sin2x cos2x

= 2 (2 sinx cosx)(1 – 2sin

^{2}x)= 2(2 × 4/5 × 3/5)(1 – 2(4/5)

^{2})= 2(24/25)(1-32/25)

= 2(24/25)((25-32)/25)

= 2(24/25)(-7/25)

= -336/625

Hence, the value of sin4x is (- 336/625)

### Question 31. If tanx = b/a, then find the value of

**Solution:**

We have to find the value of

So,

=

It is given that tanx = b/a, so

=

=

=

=

=

=

Hence, the value of is

### Question 32. If tanA = 1/7 and tanB = 1/3, show that cos2A = sin4B

**Solution:**

Given that, tanA = 1/7 and tanB = 1/3

Show: cos2A = sin4B

As we know that, tan2B = 2tanB / (1 – tan

^{2}B)= (2 × 1/3)(1 – 1/9) = 3/4

So, cos2A = (1 – tan

^{2}A)/(1 + tan^{2}A)= {1-(1/7)

^{2}}/{1+(1/7)^{2}}= 48/50

= 24/25

And sin4B = 2tan2B / (1 + tan

^{2}2B)= {2 × 3/4}{1 + (3/4)

^{2}}= 24/25

Hence, cos2A = sin4B

### Question 33. cos7° cos14° cos28° cos56° = sin68°/16cos83°

**Solution:**

Lets solve LHS

= cos7° cos14° cos28° cos56°

On dividing and multiplying by 2sin7°, we get

= × 2sin7° × cos7° × cos14° × cos28° × cos56°

= × cos28° × cos56°

= × cos56°

=

=

=

LHS = RHS

Hence proved.

### Question 34. Proved that, cos(2π/15)cos(4π/15)cos(8π/15)cos(16π/15) = 1/16

**Solution:**

Let’s solve LHS

= cos(2π/15)cos(4π/15)cos(8π/15)cos(16π/15)

On dividing and multiplying by 2sin(2π/15), we get

=

=

=

=

=

=

= 1/16

LHS = RHS

Hence proved.

### Question 35. Proved that, cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5) = -1/16

**Solution:**

Lets solve LHS

= cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5)

On dividing and multiplying by 2sin(2π/5), we get

= × 2sin(π/5)cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5)

= (sin(2π/5)cos(2π/5)cos(4π/5)cos(8π/5))

= [2sin(2π/5)cos(2π/5)cos(4π/5)cos(8π/5)]

= [sin(4π/5)cos(4π/5)cos(8π/5)]

= [2sin(4π/5)cos(4π/5)cos(8π/5)]

= [sin(8π/5)cos(8π/5)]

=[2sin(8π/5)cos(8π/5)]

=

=

=

= -1/16

LHS = RHS

Hence proved.

### Question 36. Proved that, cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65) = 1/64

**Solution:**

Lets solve LHS

= cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65)

Now on dividing and multiplying by 2sin(π/65), we get

= × 2sin(π/65)cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65)

= × [cos(2π/65) × cos(4π/65) × cos(8π/65) × cos(16π/65) × cos(32π/65)]

= × cos(4π/65) × cos(8π/65) × cos(16π/65) × cos(32π/65)

= × cos(8π/65) × cos(16π/65) × cos(32π/65)

= × cos(16π/65) × cos(32π/65)

= × cos(32π/65)

=

=

=

= 1/64

LHS = RHS

Hence proved

### Question 37. If 2tanα = 3tanβ, prove that tan(α – β) = sin2β / (5 – cos2β)

**Solution:**

Given that,

2tanα = 3tanβ

Prove: tan(α – β) = sin2β / (5 – cos2β)

Proof:

Lets solve LHS

=

=

=

=

=

=

=

=

=

=

=

=

=

=

LHS = RHS

Hence proved.

### Question 38(i). If sinα + sinβ = a and cosα + cosβ = b, prove that sin(α + β) = 2ab/(a^{2 }+ b^{2})

**Solution:**

Given that,

sinα + sinβ = a and cosα + cosβ = b

Prove: sin(α + β) = 2ab/(a

^{2 }+ b^{2})Proof:

As we know that,

So ……(i)

Now, using the identity

…..(ii)

Now on dividing eq(i) and (ii), we get

tan(α + β)/2 = a/b

As we know that,

sin2x = 2tanx/(1 + tan

^{2}x)=

= 2ab/(a

^{2 }+ b^{2})LHS = RHS

Hence proved

### Question 38(ii). If sinα + sinβ = a and cosα + cosβ = b, prove that cos(α – β) = (a^{2 }+ b^{2 }– 2)/2

**Solution:**

Given that,

sinα + sinβ = a ……(i)

cosα + cosβ = b …….(ii)

Now on squaring eq(i) and (ii) and then adding them, we get

sin

^{2}α + sin^{2}β + 2sinαsinβ + cos^{2}α + cos^{2}β + 2cosαcosβ = a^{2 }+ b^{2}⇒ 1 + 1 + 2(sinαsinβ + cosαcosβ) = a

^{2 }+ b^{2}⇒ 2(sinαsinβ + cosαcosβ) = a

^{2 }+ b^{2 }– 2⇒ 2cos(α – β) = a

^{2 }+ b^{2 }– 2⇒ cos(α – β) = (a

^{2 }+ b^{2 }– 2)/2Hence proved.

### Question 39. If 2tan(α/2) = tan(β/2), prove that cosα =

**Solution:**

Given that,

2tan(α/2) = tan(β/2)

Prove: cosα =

Proof:

Let us solve RHS

=

=

=

=

=

=

=

= cosα

RHS = LHS

Hence proved.

### Question 40. If cosx = , prove that tan(x/2) = ± tan(α/2)tan(β/2).

**Solution:**

Given that,

…..(i)

⇒

Now, by componendo and dividendo, we get

⇒

⇒

⇒

⇒

⇒ tan

^{2}(x/2) = tan^{2}(α/2)tan^{2}(β/2)⇒ tan(x/2) = ±tan(α/2)tan(β/2)

Hence Proved.

### Question 41. If sec(x + α) + sec(x – α) = 2secx, prove that cosx = ± √2 cos(α/2).

**Solution:**

Given that,

sec(x + α) + sec(x – α) = 2secx

So,

⇒

⇒

⇒

⇒ cos

^{2}xcosα = cos^{2}x(cos^{2}α + sin^{2}α) – sin^{2}α⇒ cos

^{2}x(1 – cosα) = sin^{2}α⇒

=

⇒ cosx = ± √2 cos(α/2)

Hence Proved

### Question 42. If cosα + cosβ = 1/3 and sinα + sinβ = 1/4, prove that cos(α – β)/2 = ±5/24.

**Solution:**

Given that,

cosα + cosβ = 1/3

sinα + sinβ = 1/4, we get

Prove: cos(α – β)/2 = ±5/24

Proof:

(cos

^{2}α + cos^{2}β + cosαcosβ) + (sin^{2}α + sin^{2}β + 2sinαsinβ) = 1/9 + 1/161 + 1 + 2(cosαcosβ + sinαsinβ) = 25/144

2 + 2cos(α – β) = -263/288 …..(i)

Now,

= [From (i)]

= 25/576

= ± 5/24

Hence proved.

### Question 43. If sinα = 4/5 and cosβ = 5/13, prove that cos{(α – β)/2} = 8/√65.

**Solution:**

Given that,

sinα = 4/5 and cosβ = 5/13

As we know that.

cosα = √(1 – sin

^{2}α)So,

= √{1 – (4/5)

^{2}}= 3/5

Also, sinβ = √(1 – cos

^{2}β)= √{1 – (5/13)

^{2}}= 12/13

Now,

cos(α – β) = cosα cosβ + sinα sinβ

= (3/5)(5/13)(4/5)(12/13)

= 63/65

Thus,

cos{(α – β)/2} =

=

= 8/√65

Hence Proved.

### Question 44. If acos2θ + bsin2θ = c has α and β as its roots prove that,

**(i) tanα + tanβ = 2b/(a + c)**

**(ii) tanα tanβ = (c – a)/(c + a)**

**(iii) tan(α + β) = b/a**

**Solution:**

As we know that

Now substitute these values in the given equation, we get

a(1 – tan

^{2}θ) + b(2tanθ) = c(1 + tan^{2}θ)(c + a)tan

^{2}θ + 2btanθ + c – a = 0

(i)As α and β are rootsSo, sum of the roots:

tanα + tanβ = 2b / (c + a)

(ii)As α and β are rootsSo, product of roots:

tanα tanβ = (c – a) / (c + a)

(iii)tan(α + β)==

= b/a

Hence proved.

### Question 45. If cosα + cosβ = 0 = sinα + sinβ, then prove that cos2α + cos2β = -2cos(α + β).

**Solution:**

Given that,

cosα + cosβ = 0 = sinα + sinβ

Prove: cos2α + cos2β = -2cos(α + β)

Proof:

cosα + cosβ = 0

On squaring on both sides, we get

cos

^{2}α + cos^{2}β + 2 cosα cosβ = 0 ….(i)Similarly

sinα + sinβ = 0

On squaring on both sides, we get

sin

^{2}α + sin^{2}β + 2 sinα sinβ = 0 …..(ii)Now, subtract eq (ii) from (i), we get

⇒ (cos

^{2}α + cos^{2}β + 2 cosα cosβ) – (sin^{2}α + sin^{2}β + 2 sinα sinβ) = 0⇒ cos

^{2}α – sin^{2}α + cos^{2}β – sin^{2}β + 2(cosα cosβ – sinα sinβ) = 0⇒ cos2α + cos2β + 2cos(α + β) = 0

⇒ cos2α + cos2β = -2cos(α + β)

Hence proved.